There are some very interesting subtleties here. Let's analyze the situation very carefully.
Let's choose our system to consist of the block, spring, and Earth. By choosing the Earth and block to be in our system, we will have a change in gravitational potential energy.
In the beginning, the (massless) spring hangs vertically with a block of mass $m$ attached at the bottom. We could calculate how much the spring is stretched by equating the gravitational and spring forces ($kx_1=mg$) but we won't need this.
Now, during the pulling process you describe, it's important to note that you are doing positive work on the system, which means that the energy in the system increases. It is tempting to say that the change in energy is zero, but this isn't the case for the system we've chosen.
Let's use the work-energy theorem to answer your question of where the gravitational potential energy "goes."
$$\underbrace{W_\text{net, external}}_\text{Positive}=\Delta E_\text{tot}=\underbrace{\Delta U_\text{grav}}_\text{Negative}+\underbrace{\Delta U_\text{elastic}}_\text{Positive}$$
Yes, the gravitational potential energy decreases. Where does it go? Well, the only other term that could (mathematically) compensate for this decrease in gravitational potential energy is the increase in elastic potential energy. But be careful with wording here. The spring is not storing gravitational potential energy; rather, gravitational potential energy was converted to elastic potential energy.
As a side note, since the left-hand side of the equation above is positive, the absolute value of $\Delta U_\text{elastic}$ is greater than that of $\Delta U_\text{grav}$. So, not only was the gravitational potential energy converted to elastic potential energy, the positive work you did on the system also adds to the increase in elastic potential energy.
You're missing a somewhat subtle point in your analysis. The block on the left in your diagram, where the spring is at its equilibrium position, is moving, so it has kinetic energy (which you're currently ignoring). I'll leave it to you to sort out what the speed needs to be and check that CoE holds.
It needs to be moving because, if it were not, then there must be some force (your hand?) holding it in place. This would count as an external force, and when such a force acts, CoE does not hold.
Best Answer
In SHM there is no damping so the oscillation would go on forever.In order to bring the system to equilibrium an external force would be needed and the energy stored in the spring would be diissipated to overcome this force.The energy mgx where x is the point of equilibrium to which the system has been brought by an external force equals 1/2 kx^2 (stored in the spring)+ the energy dissipated to bring the oscillation to a halt.The distance h where the energy mgh is converted totally to spring energy 1/2kh^2 is the point where the mass is at the maximum distance from equilibrium and instantaneosly at rest just before reversing direction.But this is mot equilibrium and here mgis not = kh.